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Say I have a real valued function $f(x)$, is it true that if $\lim_{x \rightarrow c} |f(x)| = 0$ then $\lim_{x \rightarrow c} f(x) = 0$?, where $c$ can be a real number or $\pm \infty$.

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Yes.

Proof. Suppose $\lim\limits_{x \to c} |f(x)| = 0$. Then $\lim\limits_{x \to c} -|f(x)| = 0$ also. For any $x$, we have $$ -|f(x)| \le f(x) \le |f(x)| $$ implying $\lim\limits_{x \to c} f(x) = 0$ by the squeeze theorem.


Note that $\lim_{x \to c} |f(x)| = L$ does not imply the existence of $\lim_{x \to c} f(x)$ when $L \ne 0$.

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  • $\begingroup$ Well, $\lim_{x \rightarrow c} |f(x)| = L$ implies that if $\lim_{x \rightarrow c} f(x)$ exists, it is either $L$ or $-L$, no? $\endgroup$
    – Lily Chung
    May 17 '14 at 20:56
  • $\begingroup$ @IstvanChung Yes, that is true. "anything about" is poor wording; I'll change it. $\endgroup$
    – 6005
    May 17 '14 at 23:42
  • $\begingroup$ I think this is not true take $f(n)=(-1)^n$. While $\lim_{n\to \infty}|f(n)|=1$ but $f(n)$ does not have a limit at infinity. $\endgroup$
    – palio
    Mar 23 '15 at 11:24
  • $\begingroup$ @palio If you are speaking about Istvan's comment, reread it. One of the hypotheses is that the limit exist. $\endgroup$
    – 6005
    Mar 24 '15 at 7:01
  • $\begingroup$ @C-S yes i reread it. Thanks !! $\endgroup$
    – palio
    Mar 25 '15 at 3:35
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Yes. Note that $\lim_{x\to c}f(x)=0$ translates to: For all $\epsilon>0$ there exists a $\delta>0$ such that for all $x$ with $0<|x-c|<\delta$ (or accordingly for the infinite case) we have $|f(x)-0|<\epsilon$. But clearly $$ |f(x)-0|<\epsilon\iff |f(x)|<\epsilon\iff \bigl||f(x)|-0\bigr|<\epsilon.$$ Hence the condition for $f$ and $|f|$ is in fact the same.

Note that for any nonzero limit $a$ we'd only have the other direction $$ |f(x)-a|<\epsilon\implies\bigl||f(x)|-|a|\bigr|<\epsilon.$$

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Yes

Try

$|f(x)|= ||f(x)|-0| < \epsilon $

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Yes, let's assume that $\lim_{x \to c} |f(x)| = 0$, then, by definition of limit:

$$\forall \varepsilon > 0, \exists \delta > 0: |x -c| < \delta \implies \big|\ |f(x)|\ \big| < \varepsilon$$

$$\iff$$

$$\forall \varepsilon > 0, \exists \delta > 0: |x -c| < \delta \implies |f(x)| < \varepsilon$$

And by the definition of limit, this means that $\lim_{x \to c} f(x) = 0$.


The proof for $x \to \infty$ ($x \to - \infty$) is almost the same.

If $\lim_{x \to \infty} |f(x)| = 0$ ($\lim_{x \to - \infty} |f(x)| = 0$) then, by the definition of limit:

$$\forall \varepsilon > 0, \exists M: x > M \implies \big|\ |f(x)|\ \big| < \varepsilon$$ $$(\forall \varepsilon > 0, \exists m: x < m \implies \big|\ |f(x)|\ \big| < \varepsilon)$$

$$\iff$$

$$\forall \varepsilon > 0, \exists M: x > M \implies |f(x)| < \varepsilon$$ $$(\forall \varepsilon > 0, \exists m: x < m \implies |f(x)| < \varepsilon)$$

So, by definition: $\lim_{x \to \infty} f(x) = 0$ ($\lim_{x \to - \infty} f(x) = 0$)

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